1. ## Points on graphs

The two exercises I'd like help with is on the picture, thanks (:

2. Hello, Fnus!

Here's the first one . . .

$f(x) \:=\:x^3 + ax + b$ has a stationary point at $(-2,3).$

(a) Find the values of $a$ and $b.$

(b) Find the position and nature of all stationary points.
The point (-2,3) is on the graph.
. . Hence: . $f(\text{-}2) = 3\!:\;\;(\text{-}2)^3 + a(\text{-}2) + b \:=\:3\quad\Rightarrow\quad \text{-}2a + b \:=\:11$ .[1]

$f'(x) \:=\:3x^2 + a$
Since (-2,3) is a stationary point, $f'(\text{-}2) = 0$
. . and we have: . $3(\text{-}2)^2 + a \:=\:0\quad\Rightarrow\quad\boxed{a \:=\:\text{-}12}$

Substitute into [1]: . $\text{-}2(\text{-}12) + b \:=\:11\quad\Rightarrow\quad\boxed{b \:=\:\text{-}13}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The function is: . $f(x) \:=\:x^3 - 12x - 13$

Then: . $f'(x) = 0\quad\Rightarrow\quad 3x^2 - 12 \:=\:0\quad\Rightarrow\quad x^2\:=\:4\quad\Rightarrow\quad x \:=\:\pm2$

When $x = 2\!:\;\;f(2) \:=\:2^3 - 12(2) - 13 \:=\:-29$ . . . Critical point: $(2,\,\text{-}29)$

When $x = \text{-}2\!:\;\;f(\text{-}2) \:=\:(\text{-}2)^3 - 12(\text{-}2) - 13 \:=\:3$ . . . Critical point: $(\text{-}2,\,3)$

Second Derivative Test: . $f''(x) \:=\:6x$

$f''(2) \:=\:6(2) \:=\:+12$ . . . concave up . . . minumum at ${\color{blue}(2, \text{-}29)}$

$f''\text{(}-2) \:=\:6(\text{-}2) \:=\:-12$ . . . concave down . . . maximum at ${\color{blue}(\text{-}2, 3)}$

3. Oh, I see.
Thanks (:
That step by step thing makes me actually understand what is going on.
Thanks a lot!

4. Here's 6

Let $P(x) = ax^3 + bx^2 + cx + d$

$\Rightarrow P'(x) = 3ax^2 + 2bx + c$

Now, let's piece our clues together.

Clue 1: $P(x)$ touches the line $y = 9x + 2$ at the point $(0,2)$

we can take two things away from this.

(1) we interpret the phrase "touching the line" to mean, the line is tangent to the curve. thus it means that $P'(0) = 2 \implies \boxed{c = 2}$

(2) the second thing we can take from this, is that $P(x)$ passes through the point $(0,2)$, that is $P(0) = 2$ which means of course, $\boxed{d = 2}$

two unknowns down, two to go.

what else do we know about $P(x)$?

Clue 2: $P(x)$ has a stationary point at $(-1,-7)$

again, there are two things we can take from this.

(1) $P(x)$ passes through $(-1,-7)$, that is, $P(-1) = -7$

that means that: $a(-1)^3 + b(-1)^2 - 2 + 2 = -7$ (i plugged in the values we got for c and d)

simplifying, we get:

$-a + b = -7$ ......................(1)

(2) the second thing we take from this is that $P'(-1) = 0$ (that is, the slope of $P(x)$ when $x = -1$ is zero)

but that means that: $3a(-1)^2 - 2b - 2 = 0$ (i plugged in the value for c)

simplifying we get:

$3a - 2b = -2$ ......................(2)

thus, to find the remaining unknowns, we must solve the system

$-a + b = -7$ ......................(1)
$3a - 2b = -2$ ......................(2)

the solutions are $\boxed{a = -16}$ and $\boxed{b = -23}$

thus, $P(x) = -16x^3 - 23x^2 + 2x + 2$

EDIT: i made an error. when i concluded that c = 2, i claimed the wrong thing, i should have said that $P'(0) = 9 \implies \boxed{c = 9}$ (since 9 is the slope of the line). make the necessary corrections, i can't bother